Why 8–15 µm is considered "thermal infrared" if typical room temperature kT is 48 µm? According to

Thaddeus Sanders

Thaddeus Sanders

Answered question

2022-05-08

Why 8–15 µm is considered "thermal infrared" if typical room temperature kT is 48 µm?
According to Wikipedia:
"Long-wavelength infrared (8–15 µm, 20–37 THz, 83–155 meV): The "thermal imaging" region, in which sensors can obtain a completely passive image of objects only slightly higher in temperature than room temperature - for example, the human body - based on thermal emissions only and requiring no illumination such as the sun, moon, or infrared illuminator. This region is also called the "thermal infrared"."
However, using h c λ = k B T, the temperature range 288–308 K (15–35 °C) is equivalent to 50–46.7 µm, while 8–15 µm is equivalent to 1800–960 K (using the same equation).

Answer & Explanation

Skyler Barber

Skyler Barber

Beginner2022-05-09Added 17 answers

Your formula is wrong, is the short answer. It works fine as an order of mangnitude approximation, but it's only an approximation. What you need is Wien's displacement law:
λ = b T
where Wien's displacement constant b = 2.8977729 ( 17 ) × 10 3 m K. Put T = 288 K , 308 K into that and you get:
λ = 9.4 10.0 μ m, which as you'd expect is at the lower range of the thermal infra-red sensors.
Note that b = h c x k B where x can be determined from the finding the peak of the black body spectrum from Planck's law - which has to be done numerically. It turns out that x = 4.96, as opposed the value of 1 you in effect used in your original estimate. Hence your values are around a factor of 5 too high.

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